Problem Statement
You are given a positive integer N. Find the number of the pairs of integers u and v(0≦u,v≦N) such that there exist two non-negative integers a and b satisfying a xorb=u and a+b=v. Here, xor denotes the bitwise exclusive OR. Since it can be extremely large, compute the answer modulo 10^9+7.
Constraints
- 1≦N≦10^{18}
Input
The input is given from Standard Input in the following format:
N
Output
Print the number of the possible pairs of integers u and v, modulo 10^9+7.
Sample Input 1
3
Sample Output 1
5
The five possible pairs of u and v are:
-
u=0,v=0 (Let a=0,b=0, then 0 xor 0=0, 0+0=0.)
-
u=0,v=2 (Let a=1,b=1, then 1 xor 1=0, 1+1=2.）
-
u=1,v=1 (Let a=1,b=0, then 1 xor 0=1, 1+0=1.）
-
u=2,v=2 (Let a=2,b=0, then 2 xor 0=2, 2+0=2.）
-
u=3,v=3 (Let a=3,b=0, then 3 xor 0=3, 3+0=3.）
Sample Input 2
1422
Sample Output 2
52277
Sample Input 3
1000000000000000000
Sample Output 3
787014179
思路：
打出前面一些项的答案：
1, 2, 4, 5, 8, 10, 13, 14, 18, 21, 26, 28, 33, 36, 40, 41, 46, 50, 57, 60, 68, 73, 80, 82, 89, 94, 102, 105, 112, 116, 121, 122, 128, 133, 142, 146, 157, 164, 174, 177, 188, 196, 209, 214, 226, 233, 242, 244, 253, 260, 272, 277, 290, 298, 309, 312, 322, 329,
可以发现这样的规律：
d
a(2k) = 2*a(k-1) + a(k);
a(2k+1) = 2*a(k) + a(k-1).
然后用数组记录前面一些项目，然后开一个map记录中间递归过程访问过的变量（即，记忆话搜索）
然后直接递归处理上面发现的递归式即可了。
这也是OEIS中的一个数列。
http://oeis.org/A007729
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细节见代码：
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <queue> #include <stack> #include <map> #include <set> #include <vector> #include <iomanip> #define ALL(x) (x).begin(), (x).end() #define rt return #define dll(x) scanf("%I64d",&x) #define xll(x) printf("%I64d ",x) #define sz(a) int(a.size()) #define all(a) a.begin(), a.end() #define rep(i,x,n) for(int i=x;i<n;i++) #define repd(i,x,n) for(int i=x;i<=n;i++) #define pii pair<int,int> #define pll pair<long long ,long long> #define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0) #define MS0(X) memset((X), 0, sizeof((X))) #define MSC0(X) memset((X), ' ', sizeof((X))) #define pb push_back #define mp make_pair #define fi first #define se second #define eps 1e-6 #define gg(x) getInt(&x) #define db(x) cout<<"== [ "<<x<<" ] =="<<endl; using namespace std; typedef long long ll; ll gcd(ll a,ll b){return b?gcd(b,a%b):a;} ll lcm(ll a,ll b){return a/gcd(a,b)*b;} ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;} inline void getInt(int* p); const int maxn=1000010; const int inf=0x3f3f3f3f; /*** TEMPLATE CODE * * STARTS HERE ***/ const ll mod=1e9+7ll; ll a[100]={1, 2, 4, 5, 8, 10, 13, 14, 18, 21, 26, 28, 33, 36, 40, 41, 46, 50, 57, 60, 68, 73, 80, 82, 89, 94, 102, 105, 112, 116, 121, 122, 128, 133, 142, 146, 157, 164, 174, 177, 188, 196, 209, 214, 226, 233, 242, 244, 253, 260, 272, 277, 290, 298, 309, 312, 322, 329, 340}; map<ll,ll> m; ll f(ll k) { // cout<<k<<endl; if(k<=50) { return a[k]; } if(m[k]) { return m[k]; } if(k&1) { return m[k]=(2ll*f(k/2ll)%mod+f(k/2ll-1ll)%mod)%mod; }else { return m[k]=(2ll*f(k/2ll-1ll)%mod+f(k/2ll)%mod)%mod; } } int main() { //freopen("D:\common_text\code_stream\in.txt","r",stdin); //freopen("D:\common_text\code_stream\out.txt","w",stdout); ll n; cin>>n; cout<<f(n)<<endl; return 0; } inline void getInt(int* p) { char ch; do { ch = getchar(); } while (ch == ' ' || ch == ' '); if (ch == '-') { *p = -(getchar() - '0'); while ((ch = getchar()) >= '0' && ch <= '9') { *p = *p * 10 - ch + '0'; } } else { *p = ch - '0'; while ((ch = getchar()) >= '0' && ch <= '9') { *p = *p * 10 + ch - '0'; } } }